1. if ab ≡ −1 (mod 24), then a + b ≡ 0 (mod 24).

2. If n ≡ −1 (mod 24), then it is not a perfect square, and the number of divisors is even.

3. If n ≡ −1 (mod 24), then 24 divides

I don't need complete proofs, just some hints would be enough.

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- Mar 22nd 2010, 01:49 AMEinStonecongruence modulo twenty four
1. if ab ≡ −1 (mod 24), then a + b ≡ 0 (mod 24).

2. If n ≡ −1 (mod 24), then it is not a perfect square, and the number of divisors is even.

3. If n ≡ −1 (mod 24), then 24 divides

I don't need complete proofs, just some hints would be enough. - Mar 22nd 2010, 02:16 AMBacterius
Hello,

(never mind what was there before)

Choose any . We have . Therefore, . Plug this result back into the second condition : . Now you might want to take a look at what the multiplicative inverses are in ;) (it is special) - Mar 22nd 2010, 02:51 AMEinStone
Thanks, that proves 1, do you have a hint for 2 as well?

- Mar 22nd 2010, 02:56 AMBacterius
- Mar 22nd 2010, 03:45 AMEinStone
Is this correct for 2.?

Assume n ≡ −1 (mod 24) is a perfect square. Then , thus c*c ≡ −1 (mod 24), by 1. we get c + c = 2c ≡ 0 (mod 24).

so 2c = M*24 => c= M*12 => , hence , a contradiction.

And how to show that there is an even number of divisors?

EDIT: Found 2.

Need proof for 3.